There is an exercise in Bourbaki about the intersection $A = k (x,y) [z] \cap k (z, x + yz)$. These are two subrings of $k (x,y,z)$. The first is PID, the second is a field. Bourbaki requests to prove that this is not a Dedekind domain.
A theorem assures us that this intersection is a Krull domain, so $A$ must be of dimension $> 1$.
When looking at the elements in $A$ we find the ring $k[z, x + yz]$.
My question is: $A = k [z, x + yz]$ ? and if not, what is it??
The answer is yes, $A = k [z, x + yz]$.
Any element $r$ of $k(z, x+ yz)$ can be written in lowest form as a quotient $\frac{g(z,x + yz)}{h(z,x + yz)}$, where $g$ and $h$ are polynomials in two variables. The fraction $\frac{g(z,x+yz)}{h(z,x+yz)}$ is also irreducible with respect to the ring $k[x,y,z] = k[z, x+ yz, y]$.
But if $r$ also belongs to $k(x,y)[z]$, then its denominator can be written in the form $f(x,y)$, where $f$ is a polynomial. After reduction to lowest form in $k[x,y,z]$, the denominator remains of the form $f_1(x,y)$, since no divisor of $f(x,y)$ can involve the variable $z$.
Thus $h(z, x+yz) = f_1(x,y)$ depends only on $x$ and $y$. Setting $y =0$, we see that $h$ cannot depend on its first variable. Setting $x = 0$, $y = 1$, we then see that $h$ cannot depend on its second variable. Thus $h$ is constant.
Therefore $r = g(z, x + yz)$